A total of 100. mL of 0.1000M HClO (Ka= 3.0x10-8)is titrated with 0.0100M KOH at 25C?
A: Calculate initial pH of the HClO solution
B: What is the pH of the solution at the equivalence point, and how many
mL of KOH solution must be added to reach the equivalence point?
A: Calculate initial pH of the HClO solution
B: What is the pH of the solution at the equivalence point, and how many
mL of KOH solution must be added to reach the equivalence point?
1 Answer
Explanation:
For which:
Since we have a weak acid we can rearrange and take -ve logs of both sides to get:
The equation is:
This tells us that 1 mole of HClO is equivalent to 1 mole of KOH.
We can find the number of moles of HClO:
The equation tells us that the no. moles of KOH must be the same:
This is the volume of KOH solution required to reach the equivalence point.
This is a badly set question as you would never set up a titration that required such a large volume. The acid is far too concentrated.
It should be diluted 10 times and only 25 ml used. That would give a sensible end - point.
We have added 0.01 moles of HClO to 0.01 moles of KOH to make 0.01 moles KClO.
We can use a corresponding expression to get the pH of a weak base:
To get
To get this we use
We know that
The new total volume = 1000ml + 100ml = 1100 ml = 1.1 L
Putting in the numbers:
Now we can use: